# Example Problem Post

This problem is from the Columbus State University problem of the week. Link

### Problem

Given $a + 1 = b + 2 = c + 3 = d + 4 = a + b + c + d + 5$, find the value of $a + b + c + d$.

### Solution

Note that this string of equalities gives us the following relations.

1. $a + 1 = b + 2$
2. $a + 1 = c + 3$
3. $a + 1= d + 4$
4. $a + 1 = a + b + c + d + 5$

We will first solve for $a$ which will in turn allow us to solve for the other values. Equations (1), (2), and (3) tell us that $b = a - 1$, $c = a -2$, and $d = a - 3$, respectively. If we substitute these values into equation (4) we find

$a + 1 = a + (a- 1) + (a-2) + (a-3) + 5$

$2 = 3a$

$a = 2/3$

Thus we can use equations (1), (2), and (3) to find that $b = -1/3$, $c = -4/3$, and $d = -7/3$. Their sum is

$a + b + c + d = 2/3 - 1/3 - 4/3 - 7/3 = -10/3$

### One response to “Example Problem Post”

1. Another way to solve this problem is using linear algebra. Note that the string of equalities gives us the following relations:
$-4 = b + c + d$
$-3 = a + c + d$
$-2 = a + b + d$
$-1 = a + b + c$
This allows us to set up two matrices:
$A = \left[\begin{array}{cccc}0 & 1 & 1 & 1\\1 & 0 & 1 & 1\\1& 1& 0 & 1\\ 1& 1& 1& 0 \end{array}\right]$
and
$b = \left[\begin{array}{c}-4 \\ -3 \\ -2 \\ -1\end{array}\right]$
Then when you compute
$A^{-1}b = \left[\begin{array}{c}2/3 \\ -1/3 \\ -4/3 \\ -7/3\end{array}\right]$
you find $a = 2/3 , b = -1/3, c = -4/3, d = -7/3$.